Problem: Is ${749628}$ divisible by $4$ ?
Explanation: A number is divisible by $4$ if the last two digits are divisible by $4$ . [ Why? We can rewrite the number as a multiple of $100$ plus the last two digits: $ \gray{7496} {28} = \gray{7496} \gray{00} + {28} $ Because $749600$ is a multiple of $100$ , it is also a multiple of $4$ So as long as the value of the last two digits, ${28}$ , is divisible by $4$ , the original number must also be divisible by $4$ Is the value of the last two digits, $28$ , divisible by $4$ Yes, ${28 \div 4 = 7}$, so $749628$ must also be divisible by $4$.